Môn Toán Lớp 8: 2x(x-3) – 3(3-x)=0 x^2 + 4x + 4=4x^2

Giải đáp:

2x ( x -3) – 3 ( 3 – x) = 0

<=> 2x ( x -3) + 3 ( x -3) = 0

<=> ( 2x +3)( x -3) = 0

<=> $\left[\begin{matrix} 2x + 3 = 0\\ x- 3 = 0\end{matrix}\right.$

<=> $\left[\begin{matrix} x= \dfrac{-3}{2}\\ x= 3\end{matrix}\right.$

Vậy S={ -3/2 ; 3}

$\\$

x^2 + 4x + 4=4x^2

<=> x^2 + 4x + 4 – 4x^2 = 0

<=> ( x^2 + 2.x.2 + 2^2) – 4x^2 = 0

<=> ( x +2)^2 – ( 2x )^2 = 0

<=> ( x +2 – 2x)( x +2 + 2x) = 0

<=> ( – x +2) ( 3x +2) = 0

<=> $\left[\begin{matrix} – x +2 = 0\\ 3x +2 = 0\end{matrix}\right.$

<=> $\left[\begin{matrix} x=2\\ x= \dfrac{-2}{3}\end{matrix}\right.$

Vậy S={ 2 ; -2/3}